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理论力学(第2版)习题答案

IT圈 admin 46浏览 0评论

2024年3月25日发(作者:隋阳德)

各章习题(计算题)部分答案

第1章 略

第2章

2-1

F

R

3.284kN

cos(F

R

,i)206.3

cos(F

R

,j)116.3

2-2

T

3

16.2kN

30

2-3

48.2

F

Rx

4.96kN

2-4

F

x1

86.6N,F

y1

50N

F

x2

30N,F

y2

40N

F

x3

0N,F

y3

60N

F

x4

56.6N,F

y4

56.6N

2-5

F

R

0

2-6(a)

F

Ax

7.07kN,F

Ay

3.54kN,F

B

3.54kN

(b)

F

Ax

0,F

Ay

5kN,F

B

5kN

(c)

F

Ax

9.33kN,F

Ay

4.33kN,F

B

6.12kN

(垂直于支撑面,指向简支梁)

2-7

F

min

15kN

F

N

25kN

2-8

F

Ax

0.866kN,F

Ay

0.5kN,T

BD

1kN

2-9

F

NA

1.732kN,F

NC

3.464kN,AC1.5m

2-10

F

AB

0,F

AC

34.36kN

2-11

F

BC

2-12

F

NE

23

3

G

F

Ax

G

F

Ay

G

3

3

6

GF

5

2-13

F

NC

F

ND

3G

2-14

F

AB

231N,F

AE

115.5N,F

BC

231N,F

BD

84.5N

PP

2-15 (a)

F

Ay

,

F

B

33

(b)

F

A

F

B

2

P

(

F

A

F

B

方向相反,组成一力偶)

3

(c)

F

A

F

B

0

777

3kN

F

AB

3kN

F

OA

3kN

F

BC

7kN

336

2-17

F

Ax

1 905N,F

Ay

1 905N,F

Cx

1 905N,F

Cy

1 905N

2-16

F

1

2-18

F

Ax

3 571N,F

Ay

3 571N,F

Cx

3 571N,F

Cy

3 571N

·312·

理论力学

2-19

M

2

4kNm

F

A

F

B

1.155kN

第3章

3-1

F

Ax

2 400N

F

Ay

1 200N

F

BC

848.5N

3-2

F'

M

O

260Nm

R

0

1

2

1

2

1

M

ql

3-3(a)

F'

(b)

M

qa

qa

F'ql

O

O

R

R

q

2

2

3-4(a)

F

Ax

(b)

F

Ax

403803

kN

F

Ay

40kN

M

A

120kNm

F

NC

kN

33

0

F

Ay

2.5kN

F

By

15kN

F

Dy

2.5kN

3-5 当

60

时,

F

ABmin

3-6

F

Ax

0

F

Ay

4Pr

L

2qa

M

A

qa

2

3-7 (a)

F

Ax

2 400N

F

Ay

1 000N

F

Dx

2 400N

F

Dy

2 000N

(b)

F

Ax

2 400N

F

Ay

1 000N

F

Dx

2 400N

F

Dy

2 000N

3-8

F

Ax



F

By

5(1

3)

14

203

4

53

kN

F

Ay

kN

F

Bx

kN

3

33

22

53

kN

3

PLPL

PL

3-9

F

Ax



F

Ay

P

F

Bx

F

By

P

F

D

F

C

2

P

2r2r

2r

3

2

qa

3-10

F

RE

qa

F

BD

2

2

3-11

F

Ax

F

Cx

2.3kN

F

Ay

F

Cy

1kN

3-12

F

AC



3-13

F

BC

第4章

4-1

F

T

20kN

F

OA

10.4kN

F

OB

13.9kN

4-2

M

x

2.54kNm

M

y

1.46kNm

M

z

0

4-3

M

z

(P)0

4-4

M

AB

(

P

)

Pasin

sin

4-5

T

A

T

B

T

C

2P

P

F

EF

0

F

BD



3

3

F

F

F

DE

2

2

W

3

4-6

T1kN

F

Ax

0

F

Ay

750N

F

Az

500N

F

Bx

433N

F

BZ

500N

4-7

F

1

F

6

F

F

3

F

F

2

F

4

F

5

0

·312

·

2024年3月25日发(作者:隋阳德)

各章习题(计算题)部分答案

第1章 略

第2章

2-1

F

R

3.284kN

cos(F

R

,i)206.3

cos(F

R

,j)116.3

2-2

T

3

16.2kN

30

2-3

48.2

F

Rx

4.96kN

2-4

F

x1

86.6N,F

y1

50N

F

x2

30N,F

y2

40N

F

x3

0N,F

y3

60N

F

x4

56.6N,F

y4

56.6N

2-5

F

R

0

2-6(a)

F

Ax

7.07kN,F

Ay

3.54kN,F

B

3.54kN

(b)

F

Ax

0,F

Ay

5kN,F

B

5kN

(c)

F

Ax

9.33kN,F

Ay

4.33kN,F

B

6.12kN

(垂直于支撑面,指向简支梁)

2-7

F

min

15kN

F

N

25kN

2-8

F

Ax

0.866kN,F

Ay

0.5kN,T

BD

1kN

2-9

F

NA

1.732kN,F

NC

3.464kN,AC1.5m

2-10

F

AB

0,F

AC

34.36kN

2-11

F

BC

2-12

F

NE

23

3

G

F

Ax

G

F

Ay

G

3

3

6

GF

5

2-13

F

NC

F

ND

3G

2-14

F

AB

231N,F

AE

115.5N,F

BC

231N,F

BD

84.5N

PP

2-15 (a)

F

Ay

,

F

B

33

(b)

F

A

F

B

2

P

(

F

A

F

B

方向相反,组成一力偶)

3

(c)

F

A

F

B

0

777

3kN

F

AB

3kN

F

OA

3kN

F

BC

7kN

336

2-17

F

Ax

1 905N,F

Ay

1 905N,F

Cx

1 905N,F

Cy

1 905N

2-16

F

1

2-18

F

Ax

3 571N,F

Ay

3 571N,F

Cx

3 571N,F

Cy

3 571N

·312·

理论力学

2-19

M

2

4kNm

F

A

F

B

1.155kN

第3章

3-1

F

Ax

2 400N

F

Ay

1 200N

F

BC

848.5N

3-2

F'

M

O

260Nm

R

0

1

2

1

2

1

M

ql

3-3(a)

F'

(b)

M

qa

qa

F'ql

O

O

R

R

q

2

2

3-4(a)

F

Ax

(b)

F

Ax

403803

kN

F

Ay

40kN

M

A

120kNm

F

NC

kN

33

0

F

Ay

2.5kN

F

By

15kN

F

Dy

2.5kN

3-5 当

60

时,

F

ABmin

3-6

F

Ax

0

F

Ay

4Pr

L

2qa

M

A

qa

2

3-7 (a)

F

Ax

2 400N

F

Ay

1 000N

F

Dx

2 400N

F

Dy

2 000N

(b)

F

Ax

2 400N

F

Ay

1 000N

F

Dx

2 400N

F

Dy

2 000N

3-8

F

Ax



F

By

5(1

3)

14

203

4

53

kN

F

Ay

kN

F

Bx

kN

3

33

22

53

kN

3

PLPL

PL

3-9

F

Ax



F

Ay

P

F

Bx

F

By

P

F

D

F

C

2

P

2r2r

2r

3

2

qa

3-10

F

RE

qa

F

BD

2

2

3-11

F

Ax

F

Cx

2.3kN

F

Ay

F

Cy

1kN

3-12

F

AC



3-13

F

BC

第4章

4-1

F

T

20kN

F

OA

10.4kN

F

OB

13.9kN

4-2

M

x

2.54kNm

M

y

1.46kNm

M

z

0

4-3

M

z

(P)0

4-4

M

AB

(

P

)

Pasin

sin

4-5

T

A

T

B

T

C

2P

P

F

EF

0

F

BD



3

3

F

F

F

DE

2

2

W

3

4-6

T1kN

F

Ax

0

F

Ay

750N

F

Az

500N

F

Bx

433N

F

BZ

500N

4-7

F

1

F

6

F

F

3

F

F

2

F

4

F

5

0

·312

·

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